2012年12月2日
Problem Statement |
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Cucumber Boy likes drawing pictures. Today, he plans to draw a picture using a very simple graphics editor.
The editor has the following functions:
At this moment, all pixels on the infinite canvas are transparent. Cucumber Boy has already stored a picture in the clipboard. You are given this picture as a vector <string> clipboard.
Cucumber Boy now wants to paste the clipboard picture onto the canvas exactly T times in a row. For each i, when pasting the clipboard for the i-th time, he will choose the pixel (i,i) as the upper left corner of the pasted picture.
You are given the vector <string> clipboard and the int T. Return the number of black pixels on the canvas after all the pasting is finished. |
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Definition |
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Constraints |
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| - | clipboard will contain between 1 and 50 elements, inclusive. | ||||||||||||
| - | Each element of clipboard will contain between 1 and 50 characters, inclusive. | ||||||||||||
| - | Each element of clipboard will contain the same number of characters. | ||||||||||||
| - | Each character of each element of clipboard will be 'B' or '.'. | ||||||||||||
| - | T will be between 1 and 1,000,000,000, inclusive. | ||||||||||||
Examples |
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只有在同一对角线的各自会影响到对方，所以按照对角线的方式来遍历每个格子，然后记录每个格子的开始和结束位置，之后做一下处理
1 #include <vector> 2 #include <string> 3 using namespace std; 4 5 class PastingPaintingDivTwo 6 { 7 public: 8 long long calc(vector<string> clipboard, int T, int x, int y) 9 { 10 long long sum = 0; 11 int endX = -1; 12 int endY = -1; 13 while(x < clipboard.size() && y < clipboard[0].size()) 14 { 15 if (clipboard[x][y] == 'B') 16 { 17 int endCurX = x + T - 1; 18 int endCurY = y + T - 1; 19 int startCurX = max(x, endX); 20 int startCurY = max(y, endY); 21 22 sum += max(0, endCurX - startCurX + 1); 23 24 endX = max(endCurX + 1, endX); 25 endY = max(endCurY + 1, endY); 26 } 27 28 x++; 29 y++; 30 } 31 32 return sum; 33 } 34 35 long long countColors(vector<string> clipboard, int T) 36 { 37 if (clipboard.size() == 0) 38 return 0; 39 40 long long sum = 0; 41 for(int i = 0; i < clipboard.size(); i++) 42 { 43 sum += calc(clipboard, T, i, 0); 44 } 45 46 for(int i = 1; i < clipboard[0].size(); i++) 47 { 48 sum += calc(clipboard, T, 0, i); 49 } 50 51 return sum; 52 } 53 };
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1
/ \
2 3
Return 6.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int calLen(TreeNode *root, int &len) 13 { 14 if (root == NULL) 15 { 16 len = 0; 17 return 0; 18 } 19 20 if (root->left == NULL && root->right == NULL) 21 { 22 len = root->val; 23 return root->val; 24 } 25 26 int leftPath, rightPath; 27 int leftLen; 28 if (root->left) 29 leftLen = calLen(root->left, leftPath); 30 else 31 { 32 leftLen = INT_MIN; 33 leftPath = 0; 34 } 35 36 int rightLen; 37 if (root->right) 38 rightLen = calLen(root->right, rightPath); 39 else 40 { 41 rightLen = INT_MIN; 42 rightPath = 0; 43 } 44 45 len = max(max(leftPath, rightPath) + root->val, root->val); 46 int maxLen = max(root->val, max(leftPath + rightPath + root->val, 47 max(leftPath + root->val, rightPath + root->val))); 48 49 return max(max(leftLen, rightLen), maxLen); 50 } 51 52 int maxPathSum(TreeNode *root) { 53 // Start typing your C/C++ solution below 54 // DO NOT write int main() function 55 int len; 56 return calLen(root, len); 57 } 58 };
2012年11月29日
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 if (prices.size() == 0) 7 return 0; 8 9 vector<int> f1(prices.size()); 10 vector<int> f2(prices.size()); 11 12 int minV = prices[0]; 13 f1[0] = 0; 14 for(int i = 1; i < prices.size(); i++) 15 { 16 minV = min(minV, prices[i]); 17 f1[i] = max(f1[i-1], prices[i] - minV); 18 } 19 20 int maxV = prices[prices.size()-1]; 21 f2[f2.size()-1] = 0; 22 for(int i = prices.size() - 2; i >= 0; i--) 23 { 24 maxV = max(prices[i], maxV); 25 f2[i] = max(f2[i+1], maxV - prices[i]); 26 } 27 28 int sum = 0; 29 for(int i = 0; i < prices.size(); i++) 30 sum = max(sum, f1[i] + f2[i]); 31 32 return sum; 33 } 34 };
2012年11月27日
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false
只能过小数据
1 class Solution { 2 public: 3 bool isMatch(const char *s, const char *p) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 if (s == NULL && p == NULL) 7 return true; 8 else if (s == NULL || p == NULL) 9 return false; 10 11 if (*p == '?') 12 { 13 if (*s == '\0') 14 return false; 15 else 16 return isMatch(s + 1, p + 1); 17 } 18 else if (*p == '*') 19 { 20 while(*s != '\0') 21 { 22 if (isMatch(s, p + 1)) 23 break; 24 s++; 25 } 26 27 if (*s != '\0') 28 return true; 29 else 30 return isMatch(s, p + 1); 31 } 32 else 33 { 34 if (*s == *p) 35 { 36 if (*s == '\0') 37 return true; 38 else 39 return isMatch(s + 1, p + 1); 40 } 41 else 42 return false; 43 } 44 } 45 };
2012年11月26日
The gray code is a binary numeral system where two successive values differ in only one bit.
Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
00 - 0 01 - 1 11 - 3 10 - 2
Note:
For a given n, a gray code sequence is not uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
1 class Solution 2 { 3 public: 4 /* 5 * for reference: http://en.wikipedia.org/wiki/Gray_code. 6 */ 7 vector<int> grayCode(int n) 8 { 9 vector<int> ret; 10 int size = 1 << n; 11 for(int i = 0; i < size; ++i) 12 ret.push_back((i >> 1)^i); 13 return ret; 14 } 15 };
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 struct Node 11 { 12 TreeNode *head; 13 TreeNode *tail; 14 Node(){} 15 Node(TreeNode *h, TreeNode *t):head(h), tail(t){} 16 }; 17 18 class Solution { 19 public: 20 Node convert(TreeNode *root) 21 { 22 if (root == NULL) 23 return Node(NULL, NULL); 24 25 Node leftNode = convert(root->left); 26 Node rightNode = convert(root->right); 27 28 root->right = NULL; 29 root->left = NULL; 30 31 if (leftNode.head) 32 { 33 root->right = leftNode.head; 34 leftNode.tail->right = rightNode.head; 35 36 } 37 else 38 { 39 root->right = rightNode.head; 40 } 41 42 TreeNode *tail; 43 if (rightNode.tail != NULL) 44 tail = rightNode.tail; 45 else if (leftNode.tail != NULL) 46 tail = leftNode.tail; 47 else 48 tail = root; 49 50 return Node(root, tail); 51 } 52 53 void flatten(TreeNode *root) { 54 // Start typing your C/C++ solution below 55 // DO NOT write int main() function 56 convert(root); 57 } 58 };
Divide two integers without using multiplication, division and mod operator.
1 class Solution { 2 private: 3 long long f[100]; 4 public: 5 int bsearch(long long a[], int left, int right, long long key) 6 { 7 if (left > right) 8 return -1; 9 10 int mid = left + (right - left) / 2; 11 if (a[mid] == key) 12 return mid; 13 else if (a[mid] < key) 14 { 15 int pos = bsearch(a, mid + 1, right, key); 16 return pos == -1 ? mid : pos; 17 } 18 else 19 { 20 return bsearch(a, left, mid - 1, key); 21 } 22 } 23 24 int divide(int dividend, int divisor) { 25 // Start typing your C/C++ solution below 26 // DO NOT write int main() function 27 int sign = dividend < 0 ? -1 : 1; 28 if (divisor < 0) 29 sign *= -1; 30 31 long long div = dividend; 32 div = abs(div); 33 long long divisorL = divisor; 34 divisorL = abs(divisorL); 35 f[0] = divisorL; 36 int size = 1; 37 while(true) 38 { 39 if (f[size-1] >= div) 40 break; 41 f[size] = f[size-1] + f[size-1]; 42 size++; 43 } 44 45 int num = 0; 46 long long sum = 0; 47 while(div > 0) 48 { 49 int pos = bsearch(f, 0, size - 1, div); 50 if (pos == -1) 51 break; 52 div -= f[pos]; 53 num += (1 << pos); 54 } 55 56 return num * sign; 57 } 58 };
Implement strStr().
Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.
1 class Solution { 2 public: 3 char *strStr(char *haystack, char *needle) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int hayLen = strlen(haystack); 7 int needLen = strlen(needle); 8 9 for(int i = 0; i <= hayLen - needLen; i++) 10 { 11 char *p = haystack + i; 12 char *q = needle; 13 while(*q != '\0') 14 { 15 if (*p != *q) 16 break; 17 else 18 { 19 p++; 20 q++; 21 } 22 } 23 24 if (*q == '\0') 25 return haystack + i; 26 } 27 28 return NULL; 29 } 30 };
题目：二叉树的结点定义如下：
struct TreeNode
{
int m_nvalue;
TreeNode* m_pLeft;
TreeNode* m_pRight;
};
输入二叉树中的两个结点，输出这两个结点在数中最低的共同父结点。
递归版本：空间O(1)，时间O(n)
1 struct Node 2 { 3 int val; 4 Node *left; 5 Node *right; 6 Node():left(NULL), right(NULL){} 7 }; 8 9 Node *findFather(Node *root, Node *node1, Node *node2, bool &node1Find, bool &node2Find) 10 { 11 if (root == NULL) 12 return NULL; 13 14 bool leftNode1Find = false, leftNode2Find = false; 15 Node *leftNode = findFather(root->left, node1, node2, leftNode1Find, leftNode2Find); 16 if (leftNode != NULL) 17 return leftNode; 18 19 bool rightNode1Find = false, rightNode2Find = false; 20 Node *rightNode = findFather(root->right, node1, node2, rightNode1Find, rightNode2Find); 21 if (rightNode != NULL) 22 return rightNode; 23 24 node1Find = leftNode1Find || rightNode1Find; 25 node2Find = leftNode2Find || rightNode2Find; 26 27 if (node1Find && node2Find) 28 return root; 29 30 if (root == node1) 31 node1Find = true; 32 33 if (root == node2) 34 node2Find = true; 35 36 return NULL; 37 }
2012年11月25日
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
For example:
Given array A = [2,3,1,1,4]
The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)
1 class Solution { 2 private: 3 int f[100000]; 4 public: 5 int jump(int A[], int n) { 6 // Start typing your C/C++ solution below 7 // DO NOT write int main() function 8 int maxPos = 0; 9 f[0] = 0; 10 11 for(int i = 0; i <= maxPos; i++) 12 { 13 int pos = i + A[i]; 14 if (pos >= n) 15 pos = n - 1; 16 17 if (pos > maxPos) 18 { 19 for(int j = maxPos + 1; j <= pos; j++) 20 f[j] = f[i] + 1; 21 maxPos = pos; 22 } 23 24 if (maxPos == n - 1) 25 return f[n-1]; 26 } 27 } 28 };